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已知[(x2+y2)-(x-y)2+2y(x-y)]÷4y=1,求4x4x

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已知[(x2+y2)-(x-y)2+2y(x-y)]÷4y=1,求
4x
4x
已知[(x2+y2)-(x-y)2+2y(x-y)]÷4y=1,求4x4x
[(x2+y2)-(x-y)2+2y(x-y)]÷4y
=(x2+y2-x2+2xy-y2+2xy-2y2)÷4y
=(4xy-2y2)÷4y
=x-
1
2y
∵x-
1
2y=1,
∴2x-y=2,

4x
4x2−y2-
1
2x+y,
=
4x
(2x+y)(2x−y)-
1
2x+y
=
4x−2x+y
(2x+y)(2x−y)
=
2x+y
(2x+y)(2x−y)
=
1
2x−y
=
1
2.